How do you solve the quadratic using the quadratic formula given #3x^2-11x=8-14x# over the set of complex numbers?

1 Answer
Aug 30, 2016

Answer:

Solution is #-1/2-sqrt105/6# and #-1/2+sqrt105/6#

Explanation:

For the equation #ax^2+bx+c=0#, the roots are given by #x=(-b+-sqrt(b^2-4ac))/(2a)#. As can be seen type of roots depend upon discriminant #b^2-4ac#. If it is negative, roots are complex and if it is a complete square (and coefficients are rational), roots are rational.

In the equation #3x^2-11x=8-14x# or #3x^2+3x-8=0#, the discriminant is #(3)^2-4×3×(-8)=9+96=105#, which is not a complete square and hence roots are irrational (not complex as mentioned in question).

Roots are #x=(-3+-sqrt105)/(2×3)#

= #(-3+-sqrt105)/6#

= #-1/2+-sqrt105/6#

Hence solution is #-1/2-sqrt105/6# and #-1/2+sqrt105/6#.